1.

E. A 4uF capacitor is charged by a 200 V supply. It is then disconnected from thesupply and connected to another uncharged 2uF capacitor. How muchelectrostatic energy of the first capacitor is lost?​

Answer»

Answer:

Capacitance of a charged CAPACITOR, C1=4μF=4×10−6F

Supply voltage, V1=200V

Electrostatic energy STORED in C1 is given by,

E1=21C1V12

      =21×4×10−6×(200)2

      =8×10−2J

Capacitance of an uncharged capacitor, C2=2μF=2×10−6F 

When C2 is connected to the circuit, the potential acquired by it is V2.

According to the conservation of CHARGE, initial charge on capacitor C1 is equal to the final charge on capacitors, C1 and C2

∴V2(C1+C2)=C1V1

V2×(4+2)×10−6=4×10−6×200

V


Discussion

No Comment Found

Related InterviewSolutions